JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Kinematics

A particle is moving along the

x

-axis with its coordinate with time '

t

' given by

x (t) = 10 + 8 t - 3 t^{2}

. Another particle is moving along the

y

-axis with its coordinate a function of time given by

y (t) = 5 - 8 t^{3}

. At

t = 1 s

, the speed of the second particle as measured in the frame of the first particle is given as

v

. Then v (in

m / s

) is
____ .

Answer: 580

Solution

Differentiating the positions, the first particle has

v_{1} = (8 - 6 t) \hat{i}

, which at

t = 1

s is

2 \hat{i}

, while the second has

v_{2} = - 24 t^{2} \hat{j}

, which at

t = 1

s is

- 24 \hat{j}

. Their relative velocity is

v_{r e l} = v_{2} - v_{1} = - 2 \hat{i} - 24 \hat{j}

, whose magnitude is

4 + 576 = 580

, so

v = 580

m/s.

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Integer answer

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